u/ruprect1047

x(9x-2a)+b(9x-2a)-(x+b)=0. In the given equation, a and b are positive constants. A solution to the equation is x=42. What is the value of a?

So I factored by grouping twice: (x+b)(9x-2a)-(x+b)=0 and then once again to get (x+b)(9x-2a-1)=0. Then I substituted 42 for x and got a to be 188.5 which I believe is the correct answer. However the question specified that both a and b are positive constants and if you were to solve for b you would get b=-42, right? This contradicts the question stem and I have never seen College Board make such a glaring mistake like this. Am I doing something wrong? I know the question is asking to solve for a but I was curious about the value of b. Thoughts?

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Can someone please explain this to me? For the last 15 years I have been using the following equation: y=A sin(Bx)+ C where A is the amplitude, B is the frequency, and C is the midline. The period is always 2Pi/B which means that the period times the frequency would always equal 2Pi.

I'm looking at this question and I know choices 1, 2, and 3 are false. I see choice 4 and I say that is false because going off the definition that I have been using forever, the frequency of this equation is 3. The answer key does indeed have choice 4 as the answer with the following explanation: Frequency is defined as the reciprocal of period. If that is the case, then frequency times period equals 1. What is going on here? According to google frequency and period are indeed inverses. How is the frequency not 3 here? If the frequency is 3/(2Pi) then I've been doing trig wrong for the last 15 years and I know that is not the case.

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So I know the answer to question is F(a) and that it is a definition of the derivative question after you integrate F'(x). However can someone explain why the following doesn't work. My first thought was to apply L'Hospital's rule since you get 0/0 when you plug in 0 for h. And then applying the FTC the derivative of the integral would just be 0, right? And then 0/1 would get me 0 which was one of the multiple choices. Where did I go wrong here?

Solve for all values of x on the interval [0,2Pi)

Sin(x)-Cos(x)= 1

So I squared both sides and ended up getting x=0, x=Pi, and x=Pi/2 but x=0 gets rejected since it is an extraneous solution. I know there is another way to do this where I write it in the form Sin(A-B). Can someone walk me through that process? Thanks in advance.